+226 There exist two complex numbers c, say c1 and c2, so that 3+2i, 6+i, and c form the vertices of an equilateral triangle. Find the product c1c2 in rectangular form.
+118687 I'd just do it like a co-ordinate geometry question
(3,2), (6,1)
\(distance=\sqrt{9+1}=\sqrt{10}\)
height from midpoint =\(\sqrt{10+2.5}=\sqrt{12.5}\)
le
gradient = 1/-3 gradient of perpendicular = 3
midpoint = (4.5, 1.5)
Equ of perpendicular
Now I would find the equation of the perpendicular bisector
And I would find the equation of the circle centre (4.5,1,5) radius sqrt12.5
and solve them simultaneously.
This is not a very 'profession' method since they are complex numbers, not co-ordinate geometry, but the method will work.
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